$\overline{AC}$ is $10$ units long $\overline{BC}$ is $9$ units long $\overline{AB}$ is $\sqrt{181}$ units long What is $\sin(\angle BAC)$ ? $A$ $C$ $B$ $10$ $9$ $\sqrt{181}$
Answer: SOH CAH TOA in = pposite over ypotenuse opposite $= \overline{BC} = 9$ hypotenuse $= \overline{AB} = \sqrt{181}$ $\sin(\angle BAC)=\frac{9}{\sqrt{181}}$ $=\dfrac{9\sqrt{181} }{181}$